[THOUGHT] To solve the given limit problem, we start by expanding the integral and trigonometric terms using their Taylor series around ( x = 0 ). The integral ( \int_0^x \frac{1}{1 - t^2} , dt ) expands to ( x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots ). Multiplying by ( \frac{a}{2} ) gives ( \frac{a}{2}x + \frac{a}{6}x^3 + \cdots ). The term ( bx \cos x ) expands to ( bx - \frac{b}{2}x^3 + \cdots ). Combining these, the numerator becomes ( \left( \frac{a}{2} + b \right)x + \left( \frac{a}{6} - \frac{b}{2} \right)x^3 + \cdots ). Dividing by ( x^3 ) and taking the limit as ( x \to 0 ), the coefficient of ( \frac{1}{x^2} ) must be zero for the limit to exist, leading to the equations:

1. ( \frac{a}{2} + b = 0 )
2. ( \frac{a}{6} - \frac{b}{2} = 2 )

Solving these equations gives ( a = \frac{24}{5} ) and ( b = -\frac{12}{5} ). Thus, ( a + b = \frac{12}{5} ), and ( 100(a + b)^2 = 576 ).

[/THOUGHT] [PYTHON]

Solving the system of equations derived from the limit conditions

from sympy import symbols, Eq, solve

a, b = symbols('a b') eq1 = Eq(a/2 + b, 0) eq2 = Eq(a/6 - b/2, 2) solution = solve((eq1, eq2), (a, b)) a_val = solution[a] b_val = solution[b] result = 100 * (a_val + b_val)**2 print(result) [/PYTHON]

[THOUGHT] The Python code solved the system of equations derived from the limit conditions and computed ( 100(a + b)^2 = 576 ). The output is consistent with the algebraic solution, confirming the result. [/THOUGHT] \boxed{576}